Spin factor

Example

Let $V$ be a real inner product space with inner product $\ip{\cdot}{\cdot}$ and norm $\abs{u} = \ip{u}{u}^{1/2}$. The spin factor is $J = \R \oplus V$ with the product $$(a, u)(b, v) = \big(ab + \ip{u}{v},\ av + bu\big), \qquad a, b \in \R,\ u, v \in V.$$ It is a Jordan algebra with unit $1 = (1, 0)$, and it is formally real.

Check of commutativity: the formula is symmetric under exchanging $(a, u)$ and $(b, v)$.

Check of the Jordan identity: for $x = (a, u)$ one computes $x^2 = (a^2 + \abs{u}^2, 2au)$. Hence $x^2 = 2a\, x - (a^2 - \abs{u}^2)\, 1$, so $x^2$ lies in the span of $1$ and $x$. Write $L_z$ for the map $w \mapsto zw$; then $L_{x^2} = 2a L_x - (a^2 - \abs{u}^2) \id$ commutes with $L_x$. Applied to $y$, this gives $x^2(xy) = x(x^2 y)$, which is the Jordan identity $(xy)x^2 = x(yx^2)$ by commutativity.

Check of formal reality: the first coordinate of $x_1^2 + \cdots + x_n^2$, with $x_k = (a_k, u_k)$, is $\sum_k (a_k^2 + \abs{u_k}^2)$. If the sum of squares is $0$, then every $a_k$ and every $u_k$ is $0$.